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Torque Help

  1. Nov 12, 2005 #1
    a.) the magnitude & direction of the resultant force(Fr) of the four forces F1,F2,F3,F4 as shown in the figure and b.) the distance from a fulcrum at pt. O to the line of action of Fr.
    Ans. Fr = 1700N
    Last edited: Nov 12, 2005
  2. jcsd
  3. Nov 12, 2005 #2
    Just so you know you can't actually show images from your harddrive unless you attach it. If you want to show an image using the [​IMG] tags then you have to host it on a website and place the url between the tags. You can use imageshack or photobucket to host images. Just google them to get the exact addresses if you want.

    Also to get helpful responses you might just want to post a couple of ideas you had on how you were thinking of approaching the question.
  4. Nov 12, 2005 #3
    Here's the figure
    My attempt:
    Fr = 0 = Fy + F1 + F3 - F4 -F2
    WAAAAAHHH help me i dont understand the problem...
  5. Nov 12, 2005 #4


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    Homework Helper

    there's NO Fy ! If the Fr was =0, would they bother to ask for it?
    (Nobody said this was a static situation)

    They want you to put Fr (=Sum of F's) out along the beam
    so that it will provide the same amount of Torque as all 4 F's it replaces.
  6. Nov 12, 2005 #5
    summation of Fy = F1 + F3 - F2 - F4
    = 3000N + 100N - 1500N - 800N

    summation of Fy= 800N

    this corect?

    Help....how do you use the distances ..... can you use torque there or
  7. Nov 12, 2005 #6


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    Homework Helper

    Equilibrium means that the Sum of Force vectors =0
    and that the Sum of Torque vectors =0.

    Would "using torque" help you to know "how to use the distances"?

    How can you make Fr provide the same amount of Torque as all 4,
    if you don't compute how much Torque these 4 Forces provide?
  8. Nov 12, 2005 #7
    but if there's a torque it will not affect the answer...
    HElp pls give me a hint
    Last edited: Nov 13, 2005
  9. Nov 13, 2005 #8
    I see what you mean. However, the torque can help you find the right force.

    In a case like this, you can't just sum up the forces. Or rather you can, but you've got to be careful abou how you do it. You found a net force of 800N, ok, but where is that force applied? You'll agree with me that it makes a difference whether you apply it close to the fulcrum or further away.

    What you have to do, is sum up the torques associated with the different forces to find the total torque.
  10. Nov 13, 2005 #9
    you can find the torque by multiplying
    the lengths and lbs ?
    How can i find the resultant force....
    Last edited: Nov 13, 2005
  11. Nov 13, 2005 #10
    let me check it step by step.
    1. What i need to find is the resultant of All these forces (F1,F3,F2, & F4)
    2. Next I will find the distance of the line of action of Fr to the fulcrum from the Pt O
    heLp can you give me a hint plsss or advice,,
    Homework helpers?
    Last edited: Nov 13, 2005
  12. Nov 13, 2005 #11
    Yes it might help me find the right force but is there a horizontal force acting
    on it?
  13. Nov 13, 2005 #12
    There is no horizontal force. All the forces you have are vertical. Why do you think there should be one?

    Did you calculate the total torque?

    Yes, that's how you do it.

    It's true that the problem is a bit confusing. I mean, with what you're given, you can certainly find the total torque, but as to the force and the distance where it's applied, they are dependent on each other. You must know one to know the other. In finding the answer you have however, I see that your book made a certain assuption which is not totally clear to me.
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